Logic Riddles for Chess Players (7)

by Arne Kaehler
8/17/2020 – The last logic riddle was really hard, and only a few readers were able to calculate all outcomes and find the correct solution. The riddle today is easier. You have to figure out how to adjust the odds in your favour. Keep in mind, you might not be able to solve the puzzle fully, but maybe your result is good enough to save the life of a grandmaster.

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The two urns

Unfortunately, a young grandmaster won a chess game against a king, with playing the King's Gambit. Not only did he mate him with a pawn, the people kiebitzing began laughing at the king. The chess player got sentenced to death by the king, right after the check-mate. Four strong guards pushed the young grandmaster into a prison cell.

The king entered the prison, carrying a sac with chess pieces. While talking to the grandmaster, he opened the sac and took out one piece after another, until only the pawns were left in the sac.

"You are quite masterful with the pieces, young grandmaster," the king said with a smirk on his face. "Especially with the pawns it seems. Let's see how much luck they bring you now, because I shall grant you one last chance to get out of here alive!" With this, he made a quick gesture and two big urns were carried in.

"Although you embarrassed me in front of my friends, I still try to be a fair King. I offer you a last chance by playing a different game than chess with me," the king said and dropped all the chess pieces out of his bag onto the floor of the cell.

"Pick up all 16 pawns and put them into the urns as you please. However, at the end both urns have to have pawns," the king said with a smile on his face, while touching both of the containers.

"After that is done, I will blindfold you, mix the containers around and you will choose one of them." The king's smile grew bigger.

"You will then pick one single pawn out of the chosen urn. If it is a white pawn, you stay alive and leave my kingdom. If you pick a black pawn, your chess career ends here."

The king continued: "I am a fair king after all. Your chances are fifty/fifty. Let us see how fate decides".

How does the grandmaster divide the pawns so that he has the greatest probability of picking a white pawn?

The king believes he is fair with giving him a 50/50 chance, but with some thinking, the grandmaster's odds to get out of the cell alive can be adjusted.

The solution of the fair fair logic riddle

If you are like me, you have no clue what the last riddle was about and how to even approach solving it. But once you understand the very basics of combinatorics, things can get very interesting. So stay with me and maybe you learn something new.

One of many helpful tutorials on YouTube about combinations and permutations

This was probably the hardest puzzle of our logic riddle series so far, and it is time to roll up all the odds and RTP's for the four games (I use ":" for "division" and "x" for "multiply"):

The ten white pawns (sorted without repetition)

Everyone who posted a comment was pretty confident, that this game was not the best to win some money over time.

The approach to solve these kinds of puzzles is to use combinatorics. 

Think about a normal poker deck which is made of 52 cards (n = 52) and the five cards in hand are described as a 5-combination (k = 5). While the cards in your hand are all distinct and the order is irrelevant, you can get 2,598,960 combinations like this. Any randomly drawn hand has a 1 / 2,598,960 chance to be picked.

Let's apply this knowledge of n and k to the games.

From the basic set of n different elements, we pick k amount of elements. Elements can only be picked once and the order of the elements is relevant.

From ten (=n) spheres, numbered from 0-9, two spheres (=k) are chosen in a particular order. With those two spheres, you can generate a number from 00 to 99, with the exceptions of 00, 11, 22, 33, 44, 55, 66, 77, 88 and of course 99. We can calculate the amount of options now. The formula is

n! : (n-k)! (the "!" stands for factorial)

In our example we have:

(1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10) : (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8) = 90

n is 10 and k is 4, leading to 5040 different outcomes.

Since the winning sum is 4000, the RTP is only 79%, thus one of the worst games to make a quick ducat with.

Black or White (sorted with repetition)

From the basic set of n different elements, we pick k amount of elements. Elements can be picked twice and the order is irrelevant.

In this example from two different containers with ten (=n) spheres each to choose from which are numbered from 0-9, and where one sphere (k=2) is chosen in no particular order.

In our example we have to calculate the amount of options with 10 (=n) spheres, numbered from 0 to 9, where one sphere (k=2) is chosen in order. Those two spheres generate a number between 00 and 99 without any exception.

The calculation requires the formula

n to the power of k which is 10(n) to the power of 2(k) = 100.

The other part is, Black and White are the two different colours on the spinning wheel (n=2) which is turned ten times (k=10), leading to a total of 1024 different results, where only one result leads to the perfect outcome.

The RTP is around 97% and very similar to the normal roulette games, but the Black king will still win some money in the end.

The Dice (Unsorted with repetition)

This is actually game four, but for obvious reasons we take a look at it before game number three.

From the basic set of n different elements, we pick k amount of elements. Elements can be picked multiple times and the order is irrelevant.

In this example from two different containers with ten (=n) spheres each to choose from which are numbered from 0-9, one sphere (k=2) is picked. With those two spheres, you can generate a number from 00 to 99 without exception, though numbers like 32 and 23 are counted as similar numbers, due to the same digits.

To calculate the amount of possible options, this formula has to be used:

(n+k-1)!:[k!(n-1)!] 

11!:(2!x9!) = 55

With n =6 and k = 4 we can reach 9!:(4!x5!) = 126 possible options by throwing the dice. But since the Black King only pays x 100 the winning amount, we end up with roughly 79%, just like the Black or White game.

Numbers of fortune (unsorted without repetition)

From the basic set of n different elements, we pick k amount of elements without being able to choose two similar Elements with the order being irrelevant.

In this example from one container with ten (=n) spheres which are numbered from 0-9, two spheres (=k) are picked.

With those two spheres, you can generate a number from 00 to 99, except for 00, 11, 22, 33, 44, 55, 66, 77, 88, and 99, and where numbers like 32 and 23 count as similar numbers, because both digits are similar.

The formula we need to calculate the options is: n! : [k!(n-k)!]

(1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10) : [(1 x 2) x (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8)] = 45

With the "Numbers of fortune" you have 10! : (4! x 6!) = 210 options.

The Black king made a big mistake! He has to pay the 250 x of the entry with an RTP of 119%.

Everybody who plays this game of "luck" long enough, will probably leave the fair fair with some ducats in their pockets.

These were a lot of formulas and numbers, and I hope I didn't make a mistake in writing them all down.

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Arne Kaehler, a creative thinker who is passionate about board games in general was born in Hamburg and learned how to play chess at a very young age. Through teaching chess to youth teams and creating chess content on YouTube, Arne was able to extend this passion onto others and has even made an online chess course for anyone who wants to learn how to play this game. Currently, Arne blogs for the English news page of ChessBase and focuses on creating promotional and entertaining articles.

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